85380
The lengths of the sub-tangent, ordinate and the sub-normal are in
1 Arithmetico geometric progressio
2 A.P.
3 H. P.
4 G.P.
Explanation:
(D) : From question, \(\because\) Subtangent \(=\mathrm{y} \times \frac{\mathrm{dx}}{\mathrm{dy}}\) and subnormal \(=\mathrm{y} \times \frac{\mathrm{dy}}{\mathrm{dx}}\) \(\therefore\) subtangent \(\times\) subnormal \(=\mathrm{y}^{2}=(\text { ordinate })^{2}\) Hence, length of the subtangent, ordinate and the subnormal are in G.P.
Karnataka CET-2012
Application of Derivatives
85381
The length of the subtangent at ' \(t\) ' on the curve \(x=a(t+\sin t), y=a(1-\cos t)\) is
1 \(2 \mathrm{a} \sin \frac{\mathrm{t}}{2}\)
2 \(2 a \sin ^{3}\left(\frac{\mathrm{t}}{2}\right) \sec \left(\frac{\mathrm{t}}{2}\right)\)
85382
The length of the subtangent to the curve \(x^{2} y^{2}=\) \(\mathbf{a}^{4}\) at \((-\mathbf{a}, \mathbf{a})\) i
1 \(a / 2\)
2 \(2 \mathrm{a}\)
3 a
4 \(a / 3\)
Explanation:
(C) : Given the curve, On differentiating w.r.t. \(\mathrm{x}\) we get, \(x^{2} \cdot 2 y \frac{d y}{d x}+y^{2} \cdot 2 x=0\) \(\frac{d y}{d x}=\frac{-y}{x}\) \(\left(\frac{d y}{d x}\right)_{(-a, a)}=-\left(\frac{a}{-a}\right)=1\) Therefore, length of subtangent at the point \((-a, a)\) \(=\frac{y}{\frac{d y}{d x}} \quad[y=a]\) \(=\frac{a}{1}=a\)
Karnataka CET-2007
Application of Derivatives
85383
The point on the curve \(y^{2}=x\), the tangent at which makes an angle \(45^{\circ}\) with \(\mathrm{X}\)-axis is
1 \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
2 \(\left(\frac{1}{2}, \frac{1}{4}\right)\)
3 \(\left(\frac{1}{2}, \frac{-1}{2}\right)\)
4 \(\left(\frac{1}{2}, \frac{1}{2}\right)\)
Explanation:
(A) : Given, The tangent makes an angle of \(45^{\circ}\) with the \(\mathrm{x}\)-axis let the required point be \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\). \(\therefore\) Slope of the tangent \(=\tan 45^{\circ}=1\) Since, The point lies on the curve, Hence, \(\mathrm{y}_{1}^{2}=\mathrm{x}_{1}\) Now, \(\quad y^{2}=x\) On differentiating both sides, \(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=1 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}}\) Slope of the tangent \(=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{1}{2 y_{1}}\) Given, \(\frac{1}{2 \mathrm{y}_{1}}=1 \Rightarrow \mathrm{y}_{1}=\frac{1}{2}\) Now, \(\mathrm{x}_{1}=\mathrm{y}_{1}^{2}\) \(\mathrm{x}_{1}=\left(\frac{1}{2}\right)^{2} \Rightarrow \mathrm{x}_{1}=\frac{1}{4}\) The point of curve, \(\left(\mathrm{x}_{1} \mathrm{y}_{1}\right)=\left(\frac{1}{4}, \frac{1}{2}\right)\)
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Application of Derivatives
85380
The lengths of the sub-tangent, ordinate and the sub-normal are in
1 Arithmetico geometric progressio
2 A.P.
3 H. P.
4 G.P.
Explanation:
(D) : From question, \(\because\) Subtangent \(=\mathrm{y} \times \frac{\mathrm{dx}}{\mathrm{dy}}\) and subnormal \(=\mathrm{y} \times \frac{\mathrm{dy}}{\mathrm{dx}}\) \(\therefore\) subtangent \(\times\) subnormal \(=\mathrm{y}^{2}=(\text { ordinate })^{2}\) Hence, length of the subtangent, ordinate and the subnormal are in G.P.
Karnataka CET-2012
Application of Derivatives
85381
The length of the subtangent at ' \(t\) ' on the curve \(x=a(t+\sin t), y=a(1-\cos t)\) is
1 \(2 \mathrm{a} \sin \frac{\mathrm{t}}{2}\)
2 \(2 a \sin ^{3}\left(\frac{\mathrm{t}}{2}\right) \sec \left(\frac{\mathrm{t}}{2}\right)\)
85382
The length of the subtangent to the curve \(x^{2} y^{2}=\) \(\mathbf{a}^{4}\) at \((-\mathbf{a}, \mathbf{a})\) i
1 \(a / 2\)
2 \(2 \mathrm{a}\)
3 a
4 \(a / 3\)
Explanation:
(C) : Given the curve, On differentiating w.r.t. \(\mathrm{x}\) we get, \(x^{2} \cdot 2 y \frac{d y}{d x}+y^{2} \cdot 2 x=0\) \(\frac{d y}{d x}=\frac{-y}{x}\) \(\left(\frac{d y}{d x}\right)_{(-a, a)}=-\left(\frac{a}{-a}\right)=1\) Therefore, length of subtangent at the point \((-a, a)\) \(=\frac{y}{\frac{d y}{d x}} \quad[y=a]\) \(=\frac{a}{1}=a\)
Karnataka CET-2007
Application of Derivatives
85383
The point on the curve \(y^{2}=x\), the tangent at which makes an angle \(45^{\circ}\) with \(\mathrm{X}\)-axis is
1 \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
2 \(\left(\frac{1}{2}, \frac{1}{4}\right)\)
3 \(\left(\frac{1}{2}, \frac{-1}{2}\right)\)
4 \(\left(\frac{1}{2}, \frac{1}{2}\right)\)
Explanation:
(A) : Given, The tangent makes an angle of \(45^{\circ}\) with the \(\mathrm{x}\)-axis let the required point be \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\). \(\therefore\) Slope of the tangent \(=\tan 45^{\circ}=1\) Since, The point lies on the curve, Hence, \(\mathrm{y}_{1}^{2}=\mathrm{x}_{1}\) Now, \(\quad y^{2}=x\) On differentiating both sides, \(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=1 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}}\) Slope of the tangent \(=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{1}{2 y_{1}}\) Given, \(\frac{1}{2 \mathrm{y}_{1}}=1 \Rightarrow \mathrm{y}_{1}=\frac{1}{2}\) Now, \(\mathrm{x}_{1}=\mathrm{y}_{1}^{2}\) \(\mathrm{x}_{1}=\left(\frac{1}{2}\right)^{2} \Rightarrow \mathrm{x}_{1}=\frac{1}{4}\) The point of curve, \(\left(\mathrm{x}_{1} \mathrm{y}_{1}\right)=\left(\frac{1}{4}, \frac{1}{2}\right)\)
85380
The lengths of the sub-tangent, ordinate and the sub-normal are in
1 Arithmetico geometric progressio
2 A.P.
3 H. P.
4 G.P.
Explanation:
(D) : From question, \(\because\) Subtangent \(=\mathrm{y} \times \frac{\mathrm{dx}}{\mathrm{dy}}\) and subnormal \(=\mathrm{y} \times \frac{\mathrm{dy}}{\mathrm{dx}}\) \(\therefore\) subtangent \(\times\) subnormal \(=\mathrm{y}^{2}=(\text { ordinate })^{2}\) Hence, length of the subtangent, ordinate and the subnormal are in G.P.
Karnataka CET-2012
Application of Derivatives
85381
The length of the subtangent at ' \(t\) ' on the curve \(x=a(t+\sin t), y=a(1-\cos t)\) is
1 \(2 \mathrm{a} \sin \frac{\mathrm{t}}{2}\)
2 \(2 a \sin ^{3}\left(\frac{\mathrm{t}}{2}\right) \sec \left(\frac{\mathrm{t}}{2}\right)\)
85382
The length of the subtangent to the curve \(x^{2} y^{2}=\) \(\mathbf{a}^{4}\) at \((-\mathbf{a}, \mathbf{a})\) i
1 \(a / 2\)
2 \(2 \mathrm{a}\)
3 a
4 \(a / 3\)
Explanation:
(C) : Given the curve, On differentiating w.r.t. \(\mathrm{x}\) we get, \(x^{2} \cdot 2 y \frac{d y}{d x}+y^{2} \cdot 2 x=0\) \(\frac{d y}{d x}=\frac{-y}{x}\) \(\left(\frac{d y}{d x}\right)_{(-a, a)}=-\left(\frac{a}{-a}\right)=1\) Therefore, length of subtangent at the point \((-a, a)\) \(=\frac{y}{\frac{d y}{d x}} \quad[y=a]\) \(=\frac{a}{1}=a\)
Karnataka CET-2007
Application of Derivatives
85383
The point on the curve \(y^{2}=x\), the tangent at which makes an angle \(45^{\circ}\) with \(\mathrm{X}\)-axis is
1 \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
2 \(\left(\frac{1}{2}, \frac{1}{4}\right)\)
3 \(\left(\frac{1}{2}, \frac{-1}{2}\right)\)
4 \(\left(\frac{1}{2}, \frac{1}{2}\right)\)
Explanation:
(A) : Given, The tangent makes an angle of \(45^{\circ}\) with the \(\mathrm{x}\)-axis let the required point be \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\). \(\therefore\) Slope of the tangent \(=\tan 45^{\circ}=1\) Since, The point lies on the curve, Hence, \(\mathrm{y}_{1}^{2}=\mathrm{x}_{1}\) Now, \(\quad y^{2}=x\) On differentiating both sides, \(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=1 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}}\) Slope of the tangent \(=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{1}{2 y_{1}}\) Given, \(\frac{1}{2 \mathrm{y}_{1}}=1 \Rightarrow \mathrm{y}_{1}=\frac{1}{2}\) Now, \(\mathrm{x}_{1}=\mathrm{y}_{1}^{2}\) \(\mathrm{x}_{1}=\left(\frac{1}{2}\right)^{2} \Rightarrow \mathrm{x}_{1}=\frac{1}{4}\) The point of curve, \(\left(\mathrm{x}_{1} \mathrm{y}_{1}\right)=\left(\frac{1}{4}, \frac{1}{2}\right)\)
85380
The lengths of the sub-tangent, ordinate and the sub-normal are in
1 Arithmetico geometric progressio
2 A.P.
3 H. P.
4 G.P.
Explanation:
(D) : From question, \(\because\) Subtangent \(=\mathrm{y} \times \frac{\mathrm{dx}}{\mathrm{dy}}\) and subnormal \(=\mathrm{y} \times \frac{\mathrm{dy}}{\mathrm{dx}}\) \(\therefore\) subtangent \(\times\) subnormal \(=\mathrm{y}^{2}=(\text { ordinate })^{2}\) Hence, length of the subtangent, ordinate and the subnormal are in G.P.
Karnataka CET-2012
Application of Derivatives
85381
The length of the subtangent at ' \(t\) ' on the curve \(x=a(t+\sin t), y=a(1-\cos t)\) is
1 \(2 \mathrm{a} \sin \frac{\mathrm{t}}{2}\)
2 \(2 a \sin ^{3}\left(\frac{\mathrm{t}}{2}\right) \sec \left(\frac{\mathrm{t}}{2}\right)\)
85382
The length of the subtangent to the curve \(x^{2} y^{2}=\) \(\mathbf{a}^{4}\) at \((-\mathbf{a}, \mathbf{a})\) i
1 \(a / 2\)
2 \(2 \mathrm{a}\)
3 a
4 \(a / 3\)
Explanation:
(C) : Given the curve, On differentiating w.r.t. \(\mathrm{x}\) we get, \(x^{2} \cdot 2 y \frac{d y}{d x}+y^{2} \cdot 2 x=0\) \(\frac{d y}{d x}=\frac{-y}{x}\) \(\left(\frac{d y}{d x}\right)_{(-a, a)}=-\left(\frac{a}{-a}\right)=1\) Therefore, length of subtangent at the point \((-a, a)\) \(=\frac{y}{\frac{d y}{d x}} \quad[y=a]\) \(=\frac{a}{1}=a\)
Karnataka CET-2007
Application of Derivatives
85383
The point on the curve \(y^{2}=x\), the tangent at which makes an angle \(45^{\circ}\) with \(\mathrm{X}\)-axis is
1 \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
2 \(\left(\frac{1}{2}, \frac{1}{4}\right)\)
3 \(\left(\frac{1}{2}, \frac{-1}{2}\right)\)
4 \(\left(\frac{1}{2}, \frac{1}{2}\right)\)
Explanation:
(A) : Given, The tangent makes an angle of \(45^{\circ}\) with the \(\mathrm{x}\)-axis let the required point be \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\). \(\therefore\) Slope of the tangent \(=\tan 45^{\circ}=1\) Since, The point lies on the curve, Hence, \(\mathrm{y}_{1}^{2}=\mathrm{x}_{1}\) Now, \(\quad y^{2}=x\) On differentiating both sides, \(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=1 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}}\) Slope of the tangent \(=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{1}{2 y_{1}}\) Given, \(\frac{1}{2 \mathrm{y}_{1}}=1 \Rightarrow \mathrm{y}_{1}=\frac{1}{2}\) Now, \(\mathrm{x}_{1}=\mathrm{y}_{1}^{2}\) \(\mathrm{x}_{1}=\left(\frac{1}{2}\right)^{2} \Rightarrow \mathrm{x}_{1}=\frac{1}{4}\) The point of curve, \(\left(\mathrm{x}_{1} \mathrm{y}_{1}\right)=\left(\frac{1}{4}, \frac{1}{2}\right)\)